A basketball is inflated in a garage at 25 degrees C to a gauge pressure of 8.0 psi. Gauge pressure is the pressure above atmospheric pressure, which is 14.7 psi. The ball is used on the driveway at a temperature of -7 degrees C and feels flat. What is the actual pressure of the air in the ball? What is the gauge pressure?
I'm pretty sure I need to use P1V1/T1= P2T2/V2 for this problem, but how can you when both the pressure and volume are unknown? Also, I don't really get the idea of gauge pressure. To find my second answer, would I add 8.0 to the first answer?
The true pressure (absolute pressure) in the ball is gauage + atmospheric. Remember if the guage pressure was zero, there is still atmospheric pressure in the ball....it is NOT empty. Use true (absolute ) pressure in these problems.
Actual pressure P1 in the ball is 8 Plus atmospheric. You are solving for P2.
You can always use that formula. In this case you do not know P2.
Change all temps to Kelvins.
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To solve this problem, you can use the ideal gas law equation, which is P1V1/T1 = P2V2/T2.
First, convert the temperatures from Celsius to Kelvin.
T1 = 25 + 273 = 298 K
T2 = -7 + 273 = 266 K
The initial pressure, P1, is the gauge pressure plus atmospheric pressure. Therefore, P1 = 8 psi + 14.7 psi = 22.7 psi.
Next, assume the volume remains constant, which means V1 = V2.
Now, you can rearrange the ideal gas law equation to solve for P2:
P1V1/T1 = P2V2/T2
Since V1 = V2, you can simplify the equation to:
P1/T1 = P2/T2
Substituting the known values:
(22.7 psi)/(298 K) = P2/(266 K)
Now, solve for P2:
P2 = (22.7 psi) * (266 K) / (298 K)
P2 ≈ 20.3 psi
Therefore, the actual pressure of the air in the ball at -7 degrees Celsius is approximately 20.3 psi.
For the gauge pressure, you can simply subtract the atmospheric pressure:
Gauge pressure = Actual pressure - Atmospheric pressure
Gauge pressure = 20.3 psi - 14.7 psi
Gauge pressure ≈ 5.6 psi
So, the gauge pressure of the air in the ball at -7 degrees Celsius is approximately 5.6 psi.
To solve this problem, you can use the ideal gas law equation, which is P1V1/T1 = P2V2/T2. Here's how you can apply it to find the actual pressure of the air in the ball:
1. Convert temperatures to Kelvin: Change the original temperature of 25 degrees Celsius to Kelvin by adding 273 (25 + 273 = 298 K). Similarly, convert the temperature of -7 degrees Celsius to Kelvin ( -7 + 273 = 266 K).
2. Identify the known values: P1 (initial pressure) is given as 8.0 psi. The initial volume of the basketball is unknown (V1), but since the basketball isn't deflated, we assume it remains constant. T1 is 298 K, and T2 is 266 K.
3. Solve for P2: Rearrange the equation P1V1/T1 = P2V2/T2 to solve for P2. Since the volume remains the same, V2/V1 = 1. Substitute the known values: (8.0 psi * 1)/(298 K) = P2/(266 K).
4. Calculate P2: Cross multiply to solve for P2. Multiply 8.0 psi by 266 K and divide the result by 298 K. P2 = (8.0 psi * 266 K) / 298 K = 7.15 psi.
So, the actual pressure of the air in the ball is approximately 7.15 psi.
Now, let's talk about gauge pressure. Gauge pressure is the pressure measured above atmospheric pressure. It represents the pressure difference compared to the pressure of the air around us. In this scenario, the gauge pressure is given as 8.0 psi. To find the absolute pressure, you need to add the atmospheric pressure, which is typically 14.7 psi at sea level.
Therefore, the absolute pressure in the ball would be the gauge pressure (8.0 psi) plus the atmospheric pressure (14.7 psi), which equals 22.7 psi.
To summarize, the actual pressure of the air in the ball is approximately 7.15 psi, and the gauge pressure is 8.0 psi. The gauge pressure does not need to be added to the actual pressure since it already includes the atmospheric pressure.